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Maximum sum rectangle in a 2D matrix

Last Updated : 02 Jul, 2025
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Given a 2D matrix mat[][] of integers, the objective is to find the submatrix (i.e., a rectangular section of the matrix) that has the maximum possible sum among all possible submatrices.

Example:

Input: mat[][] = [[1, 2, -1, -4, -20],
[-8, -3, 4, 2, 1],
[3, 8, 10, 1, 3],
[-4, -1, 1, 7, -6]]
Output: 29
Explanation: The matrix is as follows and the green rectangle denotes the maximum sum rectangle which is equal to 29.

maximum---------sum---------rectangle---------in---------a---------2d---------matrix---------3

This problem is mainly an extension of the find Largest Sum Contiguous Subarray for 1D array.  

Table of Content

[Naive approach] Iterating Over All Possible Submatrices

We explore all possible rectangles in the given 2D array, by using four variables: two to define the left and right boundaries and two more to define the top and bottom boundaries and calculate their sums, and keep track of the maximum sum found.

C++ 
#include <iostream>
#include <vector>
#include <climits>
using namespace std;

int maxRectSum(vector<vector<int>> &mat) {
  
    int n = mat.size();
    int m = mat[0].size();
    int maxSum = INT_MIN;

    for (int up = 0; up < n; up++) {
        for (int left = 0; left < m; left++) {
            for (int down = up; down < n; down++) {
                for (int right = left; right < m; right++) {
                    
                    // Find the sum of submatrix(up, right, down, left)
                    int sum = 0;
                    for (int i = up; i <= down; i++) {
                        for (int j = left; j <= right; j++) {
                            sum += mat[i][j];
                        }
                    }

                    // Update maxSum if sum > maxSum.
                    if (sum > maxSum) {
                        maxSum = sum;
                    }
                }
            }
        }
    }

    return maxSum;
}
int main() {
    vector<vector<int>> mat = {{1, 2, -1, -4, -20}, 
                               {-8, -3, 4, 2, 1}, 
                               {3, 8, 10, 1, 3}, 
                               {-4, -1, 1, 7, -6}};
                               
    cout << maxRectSum(mat) << endl;
    return 0;
}
Java 
import java.util.*;

class GfG {

    static int maxRectSum(int[][] mat) {
        int n = mat.length;
        int m = mat[0].length;
        int maxSum = Integer.MIN_VALUE;

        for (int up = 0; up < n; up++) {
            for (int left = 0; left < m; left++) {
                for (int down = up; down < n; down++) {
                    for (int right = left; right < m; right++) {
                      
                        // Find the sum of
                        // submatrix(up, right, down, left)
                        int sum = 0;
                        for (int i = up;
                             i <= down; i++) {
                            for (int j = left;
                                 j <= right; j++) {
                                sum += mat[i][j];
                            }
                        }

                        // Update maxSum if sum > maxSum.
                        if (sum > maxSum) {
                            maxSum = sum;
                        }
                    }
                }
            }
        }

        return maxSum;
    }

    public static void main(String[] args) {
        int[][] mat = { { 1, 2, -1, -4, -20 },
                        { -8, -3, 4, 2, 1 },
                        { 3, 8, 10, 1, 3 },
                        { -4, -1, 1, 7, -6 } };

        System.out.println(maxRectSum(mat));
    }
}
Python 
def maxRectSum(mat):
    n = len(mat)
    m = len(mat[0])
    maxSum = float('-inf')

    for up in range(n):
        for left in range(m):
            for down in range(up, n):
                for right in range(left, m):
                    
                    # Calculate the sum of submatrix 
                    # (up, left, down, right)
                    subMatrixSum = 0
                    for i in range(up, down + 1):
                        for j in range(left, right + 1):
                            subMatrixSum += mat[i][j]

                    # Update maxSum if a larger sum is found
                    maxSum = max(maxSum, subMatrixSum)

    return maxSum


if __name__ == "__main__":
    
    mat = [
        [1, 2, -1, -4, -20],
        [-8, -3, 4, 2, 1],
        [3, 8, 10, 1, 3],
        [-4, -1, 1, 7, -6]
    ]

    print(maxRectSum(mat))
C# 
using System;

class GfG {
    static int maxRectSum(int[][] mat) {
        int n = mat.Length;
        int m = mat[0].Length;
        int maxSum = int.MinValue;

        for (int up = 0; up < n; up++) {
            for (int left = 0; left < m; left++) {
                for (int down = up; down < n; down++) {
                    for (int right = left; right < m; right++) {
                      
                        // Calculate the sum of submatrix 
                        // (up, left, down, right)
                        int subMatrixSum = 0;
                        for (int i = up; i <= down; i++) {
                            for (int j = left; j <= right; j++) {
                                subMatrixSum += mat[i][j];
                            }
                        }

                        // Update maxSum if a larger sum is found
                        maxSum = Math.Max(maxSum, subMatrixSum);
                    }
                }
            }
        }

        return maxSum;
    }

    static void Main(string[] args) {
        int[][] mat = { 
            new int[] { 1, 2, -1, -4, -20 },
            new int[] { -8, -3, 4, 2, 1 },
            new int[] { 3, 8, 10, 1, 3 },
            new int[] { -4, -1, 1, 7, -6 } 
        };

        Console.WriteLine(maxRectSum(mat));   
    }
}
JavaScript 
function maxRectSum(mat) {
    const n = mat.length;
    const m = mat[0].length;
    let maxSum = -Infinity;

    for (let up = 0; up < n; up++) {
        for (let left = 0; left < m; left++) {
            for (let down = up; down < n; down++) {
                for (let right = left; right < m; right++) {
                
                    // Calculate the sum of submatrix 
                    // (up, left, down, right)
                    let subMatrixSum = 0;
                    for (let i = up; i <= down; i++) {
                        for (let j = left; j <= right; j++) {
                            subMatrixSum += mat[i][j];
                        }
                    }

                    // Update maxSum if a larger sum is
                    // found
                    maxSum = Math.max(maxSum, subMatrixSum);
                }
            }
        }
    }

    return maxSum;
}

// Driver Code
const mat = [
    [ 1, 2, -1, -4, -20 ], 
    [ -8, -3, 4, 2, 1 ],
    [ 3, 8, 10, 1, 3 ], 
    [ -4, -1, 1, 7, -6 ]
];

console.log(maxRectSum(mat));

Time Complexity: O((n*m)^3), as we iterate over all the boundaries of the rectangle in O((n*m) ^ 2) time. For each rectangle, we find its sum in O(n*m) time.
Auxiliary Space: O(1), constant space is used.

[Better Approach] Using Prefix Sum

In this approach, we used prefix sum matrix to efficiently compute the sum of any submatrix. We first precompute the prefix sum such that pref[i][j] stores the sum of all elements in the submatrix from the top-left corner (0, 0) to the cell (i, j). This preprocessing allows us to calculate the sum of any arbitrary submatrix in constant time using the inclusion-exclusion principle, thereby eliminating redundant computations and significantly improving performance.

Step by Step Approach:

  • Initialize a 2D prefix sum matrixpref[][] of the same size as the input matrix.
  • Precompute the prefix sum for each cell : pref[i][j] = matrix[i][j] + pref[i−1][j] + pref[i][j−1] − pref[i−1][j−1] (Handle edges separately to avoid index out-of-bound errors.)
  • Iterate over all possible top-left corners (up, left) and bottom-right corners (down, right) to define submatrices.
  • For each such submatrix, compute the sum using the prefix sum matrix in constant time: sum = pref[down][right] − pref[up - 1][right] − pref[down][left−1]+pref[up-1][left - 1]
  • Track the maximum sum obtained across all submatrices.
C++ 
#include <iostream>
#include <vector>
#include <climits>
using namespace std;

int findSum(int up, int left, int down, int right, 
            	vector<vector<int>> &pref) {
  
    // Start with the sum of the entire submatrix (0, 0) to (down, right)
    int sum = pref[down][right];
	
  	// Subtract the area to the left of the submatrix, if it exists
    if (left - 1 >= 0) {
        sum -= pref[down][left - 1];
    }

  	// Subtract the area above the submatrix, if it exists
    if (up - 1 >= 0) {
        sum -= pref[up - 1][right];
    }

    // Add back the overlapping area that was subtracted twice
    if (up - 1 >= 0 && left - 1 >= 0) {
        sum += pref[up - 1][left - 1];
    }

    return sum;
}

// function to find the maximum sum rectangle in a 2D matrix
int maxRectSum(vector<vector<int>> &mat) {
    int n = mat.size();
    int m = mat[0].size();

    // Initialize the prefix sum matrix
    vector<vector<int>> pref(n, vector<int>(m, 0));

    // Row-wise sum
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < m; j++) {
            pref[i][j] = mat[i][j];
            if (j - 1 >= 0) {
                pref[i][j] += pref[i][j - 1];
            }
        }
    }

    // Column-wise sum
    for (int j = 0; j < m; j++) {
        for (int i = 0; i < n; i++) {
            if (i - 1 >= 0) {
                pref[i][j] += pref[i - 1][j];
            }
        }
    }

    int maxSum = INT_MIN;

    for (int up = 0; up < n; up++) {
        for (int left = 0; left < m; left++) {
            for (int down = up; down < n; down++) {
                for (int right = left; right < m; right++) {
                    
                    // Find the sum of submatrix(up, right, down, left)
                    int sum = findSum(up, left, down, right, pref);

                    // Update maxSum if sum > maxSum.
                    if (sum > maxSum) {
                        maxSum = sum;
                    }
                }
            }
        }
    }

    return maxSum;
}

int main() {
    vector<vector<int>> mat = {{1, 2, -1, -4, -20}, 
                               {-8, -3, 4, 2, 1}, 
                               {3, 8, 10, 1, 3}, 
                               {-4, -1, 1, 7, -6}};
    cout << maxRectSum(mat) << endl;
}
Java 
import java.util.*;

class GfG {
 
    static int findSum(int up, int left, int down, int right,
                      int[][] pref) {
        //  Start with the sum of the entire submatrix 
        // (0, 0) to (down, right)
        int sum = pref[down][right];
    	
      	// Subtract the area to the left of the submatrix, if it exists
        if (left - 1 >= 0) {
            sum -= pref[down][left - 1];
        }
    
      	// Subtract the area above the submatrix, if it exists
        if (up - 1 >= 0) {
            sum -= pref[up - 1][right];
        }
    
        // Add back the overlapping area that was subtracted twice
        if (up - 1 >= 0 && left - 1 >= 0) {
            sum += pref[up - 1][left - 1];
        }
    
        return sum;
    }

    // function to find the maximum sum rectangle in a 2D matrix
    public static int maxRectSum(int[][] mat) {
        int n = mat.length;
        int m = mat[0].length;

        // Initialize the prefix sum matrix
        int[][] pref = new int[n][m];

        // Row-wise sum
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                pref[i][j] = mat[i][j];
                if (j - 1 >= 0) {
                    pref[i][j] += pref[i][j - 1];
                }
            }
        }

        // Column-wise sum
        for (int j = 0; j < m; j++) {
            for (int i = 0; i < n; i++) {
                if (i - 1 >= 0) {
                    pref[i][j] += pref[i - 1][j];
                }
            }
        }

        // Find the maximum sum rectangle
        int maxSum = Integer.MIN_VALUE;

        for (int up = 0; up < n; up++) {
            for (int left = 0; left < m; left++) {
                for (int down = up; down < n; down++) {
                    for (int right = left; right < m;
                         					right++) {
                        // Find the sum of the submatrix
                        // (up, left) to
                        // (down, right)
                        int sum = findSum( up, left, down, 
                                          	right, pref);

                        // Update maxSum if sum > maxSum
                        if (sum > maxSum) {
                            maxSum = sum;
                        }
                    }
                }
            }
        }

        return maxSum;
    }

    public static void main(String[] args) {
        int[][] mat = { { 1, 2, -1, -4, -20 },
                        { -8, -3, 4, 2, 1 },
                        { 3, 8, 10, 1, 3 },
                        { -4, -1, 1, 7, -6 } };

        System.out.println(maxRectSum(mat));
    }
}
Python 
def findSum(up, left, down, right, pref):
    # Start with the sum of the entire submatrix (0, 0) to (down, right)
    totalSum = pref[down][right]

    # Subtract the area to the left of the submatrix, if it exists
    if left - 1 >= 0:
        totalSum -= pref[down][left - 1]

    # Subtract the area above the submatrix, if it exists
    if up - 1 >= 0:
        totalSum -= pref[up - 1][right]
    
    # Add back the overlapping area that was subtracted twice
    if up - 1 >= 0 and left - 1 >= 0:
        totalSum += pref[up - 1][left - 1]

    return totalSum

# Function to find the maximum sum rectangle in a 2D matrix
def maxRectSum(mat):
    n = len(mat)
    m = len(mat[0])

    # Initialize the prefix sum matrix
    pref = [[0] * m for _ in range(n)]

    # Row-wise sum
    for i in range(n):
        for j in range(m):
            pref[i][j] = mat[i][j]
            if j - 1 >= 0:
                pref[i][j] += pref[i][j - 1]

    # Column-wise sum
    for j in range(m):
        for i in range(n):
            if i - 1 >= 0:
                pref[i][j] += pref[i - 1][j]

    # Find the maximum sum rectangle
    maxSum = float('-inf')

    for up in range(n):
        for left in range(m):
            for down in range(up, n):
                for right in range(left, m):
                    totalSum = findSum(up, left, down, right, pref)
                    maxSum = max(maxSum, totalSum)

    return maxSum

if __name__ == "__main__":
    mat = [
        [1, 2, -1, -4, -20],
        [-8, -3, 4, 2, 1],
        [3, 8, 10, 1, 3],
        [-4, -1, 1, 7, -6]
    ]

    print(maxRectSum(mat))
C# 
using System;

class GfG {
    static int findSum(int up, int left, int down, int right, int[,] pref) {
        // Start with the sum of the entire submatrix from (0,0) to (down,right)
        int sum = pref[down, right];

        // Subtract the area to the left of the submatrix, if it exists
        if (left - 1 >= 0)
            sum -= pref[down, left - 1];

        // Subtract the area above the submatrix, if it exists
        if (up - 1 >= 0)
            sum -= pref[up - 1, right];

        // Add back the overlapping area that was subtracted twice
        if (up - 1 >= 0 && left - 1 >= 0)
            sum += pref[up - 1, left - 1];

        return sum;
    }

    static int maxRectSum(int[,] mat) {
        int n = mat.GetLength(0);
        int m = mat.GetLength(1);

        // Initialize the prefix sum matrix
        int[,] pref = new int[n, m];

        // Compute row-wise prefix sum
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                pref[i, j] = mat[i, j];
                if (j - 1 >= 0)
                    pref[i, j] += pref[i, j - 1];
            }
        }

        // Compute column-wise prefix sum
        for (int j = 0; j < m; j++) {
            for (int i = 0; i < n; i++) {
                if (i - 1 >= 0)
                    pref[i, j] += pref[i - 1, j];
            }
        }

        int maxSum = int.MinValue;

        // Check all possible submatrices
        for (int up = 0; up < n; up++) {
            for (int left = 0; left < m; left++) {
                for (int down = up; down < n; down++) {
                    for (int right = left; right < m; right++) {
                        int totalSum = findSum(up, left, down, right, pref);
                        if (totalSum > maxSum)
                            maxSum = totalSum;
                    }
                }
            }
        }

        return maxSum;
    }

    static void Main(string[] args) {
        int[,] mat = new int[,]
        {
            { 1, 2, -1, -4, -20 },
            { -8, -3, 4, 2, 1 },
            { 3, 8, 10, 1, 3 },
            { -4, -1, 1, 7, -6 }
        };

        Console.WriteLine(maxRectSum(mat)); 
    }
}
JavaScript 
// Function to get the sum of submatrix (r1, c1) to (r2, c2)
function findSum(up, left, down, right, pref) {
    
    // Start with the sum of the entire submatrix (0, 0) to (down, right)
    let sum = pref[down][right];
	
  	// Subtract the area to the left of the submatrix, if it exists
    if (left - 1 >= 0) {
        sum -= pref[down][left - 1];
    }

  	// Subtract the area above the submatrix, if it exists
    if (up - 1 >= 0) {
        sum -= pref[up - 1][right];
    }

    // Add back the overlapping area that was subtracted twice
    if (up - 1 >= 0 && left - 1 >= 0) {
        sum += pref[up - 1][left - 1];
    }

    return sum;
}

// function to find the maximum sum rectangle in a 2D matrix
function maxRectSum(mat) {
    const n = mat.length;
    const m = mat[0].length;

    // Create and initialize prefix sum matrix
    const pref = Array.from({ length: n }, () => Array(m).fill(0));

    // Row-wise prefix sum
    for (let i = 0; i < n; i++) {
        for (let j = 0; j < m; j++) {
            pref[i][j] = mat[i][j];
            if (j - 1 >= 0) {
                pref[i][j] += pref[i][j - 1];
            }
        }
    }

    // Column-wise prefix sum
    for (let j = 0; j < m; j++) {
        for (let i = 0; i < n; i++) {
            if (i - 1 >= 0) {
                pref[i][j] += pref[i - 1][j];
            }
        }
    }

    let maxSum = -Infinity;

    // Try all submatrices
    for (let up = 0; up < n; up++) {
        for (let left = 0; left < m; left++) {
            for (let down = up; down < n; down++) {
                for (let right = left; right < m; right++) {
                    let sum = findSum(up, left, down, right, pref);
                    if (sum > maxSum) {
                        maxSum = sum;
                    }
                }
            }
        }
    }

    return maxSum;
}

// Driver code
const mat = [
    [ 1, 2, -1, -4, -20 ],
    [ -8, -3, 4, 2, 1 ],
    [ 3, 8, 10, 1, 3 ],
    [ -4, -1, 1, 7, -6 ]
];

console.log(maxRectSum(mat));  // Output: 29

Time Complexity: O((n*m)^2), due to iteration over all the boundaries of the rectangle.
Auxiliary Space: O(n*m) due to the prefix sum matrix.

[Expected Approach] Using Kadane’s Algorithm

Instead of checking every possible submatrix, we fix two column boundaries: left and right. For each pair of columns, we compress the 2D matrix into a 1D array, where each element represents the sum of rows between the current left and right columns. This 1D array (temp[]) represents the column-wise sum for a fixed column window.
Now, we can apply Kadane’s algorithm on this temp[] array to find the maximum subarray sum, which corresponds to a rectangular submatrix with the maximum sum between columns left and right.

C++ 
#include <iostream>
#include <vector>
#include <climits>

using namespace std;

// Kadane's algorithm to find the maximum sum 
// subarray in a 1D array
int kadane(vector<int>& temp) {
  	int rows = temp.size();
    int currSum = 0;
    int maxSum = INT_MIN;

    for (int i = 0; i < rows; i++) {
        currSum += temp[i];

        // Update maxSum if the current sum is greater
        if (maxSum < currSum) {
            maxSum = currSum;
        }

        // If the current sum becomes negative, reset it to 0
        if (currSum < 0) {
            currSum = 0;
        }
    }

    return maxSum;
}

// Function to find the maximum sum rectangle in a 2D matrix
int maxRectSum(vector<vector<int>> &mat) {
    int rows = mat.size();
    int cols = mat[0].size();

    int maxSum = INT_MIN;

    // Initialize a temporary array to store row-wise
  	// sums between left and right boundaries
    vector<int> temp(rows);

    // Check for all possible left and right boundaries
    for (int left = 0; left < cols; left++) {
      
        // Reset the temporary array for each new left boundary
        for (int i = 0; i < rows; i++)
            temp[i] = 0;

        for (int right = left; right < cols; right++) {
          
            // Update the temporary array with the current
          	// column's values
            for (int row = 0; row < rows; row++) {
                temp[row] += mat[row][right];
            }

            // Find the maximum sum of the subarray for the 
          	// current column boundaries
            int sum = kadane(temp);

            // Update the maximum sum found so far
            maxSum = max(maxSum, sum);
        }
    }

    return maxSum;
}

int main() {
    vector<vector<int>> mat = {{1, 2, -1, -4, -20},
                               {-8, -3, 4, 2, 1}, 
                               {3, 8, 10, 1, 3},
                               {-4, -1, 1, 7, -6}};

    int res = maxRectSum(mat); 
    cout << res << endl;             

    return 0;
}
Java 
class GfG {

    // Kadane's algorithm to find the maximum sum subarray
    // in a 1D array
    static int kadane(int[] temp) {
        int rows = temp.length;
        int currSum = 0;
        int maxSum = Integer.MIN_VALUE;

        for (int i = 0; i < rows; i++) {
            currSum += temp[i];

            // Update maxSum if the current sum is greater
            if (maxSum < currSum) {
                maxSum = currSum;
            }

            // If the current sum becomes negative, reset it
            // to 0
            if (currSum < 0) {
                currSum = 0;
            }
        }

        return maxSum;
    }

    // Function to find the maximum sum rectangle in a 2D
    // matrix
    static int maxRectSum(int[][] mat) {
        int rows = mat.length;
        int cols = mat[0].length;

        int maxSum = Integer.MIN_VALUE;

        // Initialize a temporary array to store row-wise
        // sums between left and right boundaries
        int[] temp = new int[rows];

        // Check for all possible left and right boundaries
        for (int left = 0; left < cols; left++) {
          
            // Reset the temporary array for each new left
            // boundary
            for (int i = 0; i < rows; i++) {
                temp[i] = 0;
            }

            for (int right = left; right < cols; right++) {
              
                // Update the temporary array with the
                // current column's values
                for (int row = 0; row < rows; row++) {
                    temp[row] += mat[row][right];
                }

                // Find the maximum sum of the subarray for
                // the current column boundaries
                int sum = kadane(temp);

                // Update the maximum sum found so far
                maxSum = Math.max(maxSum, sum);
            }
        }

        return maxSum;
    }

    public static void main(String[] args) {
        int[][] mat = { { 1, 2, -1, -4, -20 },
                        { -8, -3, 4, 2, 1 },
                        { 3, 8, 10, 1, 3 },
                        { -4, -1, 1, 7, -6 } };

        int res = maxRectSum(mat);
        System.out.println(res);
    }
}
Python 
def kadane(temp):
    rows = len(temp)

    currSum = 0
    maxSum = float('-inf')

    for i in range(rows):
        currSum += temp[i]

        # Update maxSum if the current sum is greater
        if maxSum < currSum:
            maxSum = currSum

        # If the current sum becomes negative, reset it to 0
        if currSum < 0:
            currSum = 0

    return maxSum


def maxRectSum(mat):
    rows = len(mat)
    cols = len(mat[0])

    maxSum = float('-inf')

    # Initialize a temporary array to store row-wise
    # sums between left and right boundaries
    temp = [0] * rows

    # Check for all possible left and right boundaries
    for left in range(cols):
        # Reset the temporary array for each new left
        # boundary
        temp = [0] * rows

        for right in range(left, cols):
            # Update the temporary array with the current
            # column's values
            for row in range(rows):
                temp[row] += mat[row][right]

            # Find the maximum sum of the subarray for the
            # current column boundaries
            sumValue = kadane(temp)

            # Update the maximum sum found so far
            maxSum = max(maxSum, sumValue)

    return maxSum

if __name__ == "__main__":
    
    mat = [
        [1, 2, -1, -4, -20],
        [-8, -3, 4, 2, 1],
        [3, 8, 10, 1, 3],
        [-4, -1, 1, 7, -6]
    ]

    res = maxRectSum(mat)
    print(res)
C# 
using System;

class GfG {
  
    // Function to apply Kadane's algorithm to find the
    // maximum sum subarray
    static int kadane(int[] temp) {
        int rows = temp.Length;
        int currSum = 0;
        int maxSum = int.MinValue;

        for (int i = 0; i < rows; i++) {
            currSum += temp[i];

            // Update maxSum if the current sum is greater
            if (maxSum < currSum) {
                maxSum = currSum;
            }

            // If the current sum becomes negative, reset it
            // to 0
            if (currSum < 0) {
                currSum = 0;
            }
        }

        return maxSum;
    }

    // Function to find the maximum sum of submatrix
    static int maxRectSum(int[, ] mat) {
        int rows = mat.GetLength(0);
        int cols = mat.GetLength(1);

        int maxSum = int.MinValue;

        // Initialize a temporary array to store row-wise
        // sums between left and right boundaries
        int[] temp = new int[rows];

        // Check for all possible left and right boundaries
        for (int left = 0; left < cols; left++) {
          
            // Reset the temporary array for each new left
            // boundary
            Array.Clear(temp, 0, rows);

            for (int right = left; right < cols; right++) {
              
                // Update the temporary array with the
                // current column's values
                for (int row = 0; row < rows; row++) {
                    temp[row] += mat[row, right];
                }

                // Find the maximum sum of the subarray for
                // the current column boundaries
                int sumValue = kadane(temp);

                // Update the maximum sum found so far
                maxSum = Math.Max(maxSum, sumValue);
            }
        }

        return maxSum;
    }

    static void Main() {
        
        int[, ] mat = { { 1, 2, -1, -4, -20 },
                        { -8, -3, 4, 2, 1 },
                        { 3, 8, 10, 1, 3 },
                        { -4, -1, 1, 7, -6 } };

        int res = maxRectSum(mat);
        Console.WriteLine(res);
    }
}
JavaScript 
// Function to apply Kadane's algorithm to find the maximum
// sum subarray
function kadane(temp) {
    let rows = temp.length;
    let currSum = 0;
    let maxSum = -Infinity;

    for (let i = 0; i < rows; i++) {
        currSum += temp[i];

        // Update maxSum if the current sum is greater
        if (maxSum < currSum) {
            maxSum = currSum;
        }

        // If the current sum becomes negative, reset it to
        // 0
        if (currSum < 0) {
            currSum = 0;
        }
    }

    return maxSum;
}

// Function to find the maximum sum of submatrix
function maxRectSum(mat) {
    const rows = mat.length;
    const cols = mat[0].length;

    let maxSum = -Infinity;

    // Initialize a temporary array to store row-wise sums
    // between left and right boundaries
    let temp = new Array(rows).fill(0);

    // Check for all possible left and right boundaries
    for (let left = 0; left < cols; left++) {
    
        // Reset the temporary array for each new left
        // boundary
        temp.fill(0);

        for (let right = left; right < cols; right++) {
        
            // Update the temporary array with the current
            // column's values
            for (let row = 0; row < rows; row++) {
                temp[row] += mat[row][right];
            }

            // Find the maximum sum of the subarray for the
            // current column boundaries
            let sumValue = kadane(temp, rows);

            // Update the maximum sum found so far
            maxSum = Math.max(maxSum, sumValue);
        }
    }

    return maxSum;
}

// Driver Code
const mat = [
    [ 1, 2, -1, -4, -20 ], 
    [ -8, -3, 4, 2, 1 ],
    [ 3, 8, 10, 1, 3 ], 
    [ -4, -1, 1, 7, -6 ]
];

const res = maxRectSum(mat);
console.log(res);

Time complexity: O(n*m^2), for each pair of column boundaries, we compute a compressed row sum and apply Kadane’s algorithm.
Auxiliary Space: O(n) we used for the temporary array to store cumulative row sums.


Next Article
Maximum sum rectangle in a 2D matrix

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